Q. If $f(x) = \int_{- 1}^{x}{|t|dt,}$then for any $x \geq 0,f(x)$is equal to
A) $1 - x^{2}$
B) $\frac{1}{2}(1 + x^{2})$
C) $1 + x^{2}$
D) $\frac{1}{2}(1 - x^{2})$
Explanation
(b)
Given, $f(x) = \int_{- 1}^{x}{|t|dt}$
$= \int_{- 1}^{0}{|t|dt + \int_{0}^{x}{|t|dt}}$
$= \int_{- 1}^{0}{- t\ dt + \int_{0}^{x}{t\ dt}}$
$= - \left\lbrack \frac{t^{2}}{2} \right\rbrack_{- 1}^{0} + \left\lbrack \frac{t^{2}}{2} \right\rbrack_{0}^{x}$
$= - \left\lbrack 0 - \frac{1}{2} \right\rbrack + \left\lbrack \frac{x^{2}}{2} - 0 \right\rbrack = \frac{1}{2}(1 + x^{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.